Question

Let $Y\left(\omega \right)$ is the fourier transform of the signal $y\left(t\right)=20x\left(5t\right)$ where $x\left(t\right)$ is shown as

Then the value of $Y\left(\omega \right)$ at $\omega$ = 20 is equal to ________.

• A. 1.82

Answer 1

The correct option is A 1.82
Given, $y\left(t\right)=20x\left(5t\right)$
By taking Fourier transform,
$Y\left(\omega \right)=20\times \frac{1}{5}X\left(\frac{\omega }{5}\right)=4X\left(\frac{\omega }{5}\right)$
$\therefore x\left(at\right)\stackrel{FT}{⟷}\frac{1}{|a|}X\left(\frac{\omega }{a}\right)$
but $x\left(t\right)=rect\left(t\right)$
$\therefore X\left(\omega \right)=sinc\left(\frac{\omega }{2}\right)=\frac{sin\left(\frac{\omega }{2}\right)}{\frac{\omega }{2}}$
$\therefore Y\left(\omega \right)=4\frac{sin\left(\frac{\omega }{10}\right)}{\left(\frac{\omega }{10}\right)}$

$At\omega =20rad/sec,Y\left(20\right)=4\frac{sin\left(2\right)}{2}=2sin\left(2\right)$

Here in sin 2, 2 is in radians, so 2 radians = 114.59 degrees

$\therefore Y\left(20\right)=1.82$